Question

Jason and Jenny are out on a lunch date. Jason gives Jenny a bouquet that has 4 roses, 3 carnations, and 4 orchids. They decide that Jenny gets to choose their lunch order if the flower she randomly picks from the bouquet is a carnation or a rose. What is the probability that Jenny gets to choose lunch?

Answer 1

Answer: The probability that Jenny gets to choose lunch =$\frac{7}{11}$

Step-by-step explanation:

Choosing any color flower from the bouquet makes events mutually exclusive.

Now, the total number of flowers in the bouquet= 4+3+4 = 11

Number of rose=4

Probability of choosing a rose P(R)=$\frac{4}{11}$

Number of carnations=3

Probability of choosing a carnation P(C)=$\frac{3}{11}$

The probability of choosing a rose or a carnation

P(R or C)=P(R)+P(C)=$\frac{4}{11}+\frac{3}{11}=\frac{7}{11}$

Thus, the probability that Jenny gets to choose lunch =$\frac{7}{11}$

Answer 2

Answer:

7/11

Step-by-step explanation:

4+3+4=11 and 7 of the flowers are either roses or carnations.